Signal-signal beat interference (SSBI) causes performance degradation of direct detection orthogonal frequency division multiplexing (OFDM) systems. Assuming an optical complex modulator is used, the optical field E at the modulator output consists of a dc carrier and a digital signal s:E(n)=a0+s(n)
The electrical signal after direct detection, y, can be represented byy(n)=|E(n)|2=|a0|2+a0·s*(n)+a0*·s(n)+|s(n)|2 
The last term represents SSBI. When dealing specifically with an OFDM signal, the optical field can be represented as
      E    ⁡          (      n      )        =                    a        0            +              s        ⁡                  (          n          )                      =                  a        0            +                        ∑                      k            =                                          -                N                            ⁢                              /                            ⁢              2                                            N            ⁢                          /                        ⁢            2                          ⁢                                  ⁢                              a            k                    ·                      exp            ⁡                          (                              j                ⁢                                                                  ⁢                2                ⁢                                                                  ⁢                π                ⁢                                                                  ⁢                nk                ⁢                                  /                                ⁢                N                            )                                          
The electrical signal after direct detection can be represented as
      y    ⁡          (      n      )        =                                      E          ⁡                      (            n            )                                      2        =                                                  a            0                                    2            +                        a          0          *                ⁢                              ∑                          k              =                                                -                  N                                ⁢                                  /                                ⁢                2                                                    N              ⁢                              /                            ⁢              2                                ⁢                                          ⁢                                    a              k                        ·                          exp              ⁡                              (                                  j                  ⁢                                                                          ⁢                  2                  ⁢                                                                          ⁢                  π                  ⁢                                                                          ⁢                  nk                  ⁢                                      /                                    ⁢                  N                                )                                                        +                        a          0                ⁢                              ∑                          k              =                                                -                  N                                ⁢                                  /                                ⁢                2                                                    N              ⁢                              /                            ⁢              2                                ⁢                                    a              k              *                        ·                          exp              ⁡                              (                                                      -                    j                                    ⁢                                                                          ⁢                  2                  ⁢                                                                          ⁢                  π                  ⁢                                                                          ⁢                  nk                  ⁢                                      /                                    ⁢                  N                                )                                                        +                        ∑                      k            =                                          -                N                            ⁢                              /                            ⁢              2                                            N            ⁢                          /                        ⁢            2                          ⁢                                            a              k                        ·                          exp              ⁡                              (                                  j                  ⁢                                                                          ⁢                  2                  ⁢                                                                          ⁢                  π                  ⁢                                                                          ⁢                  nk                  ⁢                                      /                                    ⁢                  N                                )                                              ⁢                                    ∑                              k                =                                                      -                    N                                    ⁢                                      /                                    ⁢                  2                                                            N                ⁢                                  /                                ⁢                2                                      ⁢                                          a                k                *                            ·                              exp                ⁡                                  (                                                            -                      j                                        ⁢                                                                                  ⁢                    2                    ⁢                                                                                  ⁢                    π                    ⁢                                                                                  ⁢                    nk                    ⁢                                          /                                        ⁢                    N                                    )                                                                        
The last term in the equation above is SSBI.
Carrier-signal power ratio (CSPR) is a key parameter in a direct detection OFDM system. FIG. 1 is a graph 100 that shows a typical bit error rate (BER) and CSPR relationship. When CSPR is small, BER is worse because SSBI power is increased relative to signal power, whereas when CSPR is large, BER is also worse because system noise (usually approximated by Gaussian) dominates. The sweet spot for CPSR varies with system noise. In this illustration, system noise only consists of optical noise and is defined by optical signal-to-noise ratio (OSNR). Unlike the Gaussian nature of system noise, SSBI is deterministic relative to the signal itself. This deterministic nature can be exploited to improve error performance in an SSBI limited regime.
One method reserves a gap between the dc carrier and OFDM signal. As a result, SSBI falls into the gap after direct detection and its distortion to signal becomes smaller. Another method is to use only the odd subcarriers to carry data symbols. Therefore SSBI falls on the even subcarriers and does not cause interference to the signal itself. FIG. 2 is a diagram 200 illustrating these two schemes. An obvious drawback of these two schemes is reduced spectrum efficiency, so larger bandwidth hardware and higher sampling ADC/DAC are required.
Another method is to perform iterative SSBI cancellation at the RX side as shown in FIG. 3, which illustrates a flowchart for a SSBI cancellation method 300 at the RX side. SSBI is reconstructed from decision symbols and subtracted from the received electrical signal. Because of decision error, multiple iterations are often needed to achieve sizeable gain, which renders this method more computational complex.